Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $47.2$ years; the standard deviation is $4.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $61$ years.
$47.2$ $42.6$ $51.8$ $38$ $56.4$ $33.4$ $61$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $47.2$ years. We know the standard deviation is $4.6$ years, so one standard deviation below the mean is $42.6$ years and one standard deviation above the mean is $51.8$ years. Two standard deviations below the mean is $38$ years and two standard deviations above the mean is $56.4$ years. Three standard deviations below the mean is $33.4$ years and three standard deviations above the mean is $61$ years. We are interested in the probability of a bear living less than $61$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $33.4$ years and the other half $({0.15\%})$ will live longer than $61$ years. The probability of a particular bear living less than $61$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.